A016209 - OEIS

%I #51 Oct 31 2023 10:58:07

%S 1,9,58,330,1771,9219,47188,239220,1205941,6059229,30384718,152189310,

%T 761743711,3811110039,19062724648,95335146600,476740303081,

%U 2383895225649,11920057258978,59602029687090

%N Expansion of 1/((1-x)(1-3x)(1-5x)).

%C For a combinatorial interpretation following from a(n) = A039755(n+2,2) = h^{(3)}_n, the complete homogeneous symmetric function of degree n in the symbols {1, 3, 5} see A039755. - _Wolfdieter Lang_, May 26 2017

%H Vincenzo Librandi, <a href="/A016209/b016209.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-23,15).

%F a(n) = A039755(n+2, 2).

%F a(n) = (5^(n+2) - 2*3^(n+2)+1)/8 = a(n-1) + A005059(n+1) = 8*a(n-1) - 15*a(n-2) + 1 = (A003463(n+2) - A003462(n+2))/2. - _Henry Bottomley_, Jun 06 2000

%F G.f.: 1/((1-x)(1-3*x)(1-5*x)). See the name.

%F E.g.f.: (25*exp(5*x) - 18*exp(3*x) + exp(x))/8, from the e.g.f. of the third column (k=2) of A039755. - _Wolfdieter Lang_, May 26 2017

%e a(2) = h^{(3)}_2 = 1^2 + 3^2 + 5^2 + 1^1*(3^1 + 5^1) + 3^1*5^1 = 58. - _Wolfdieter Lang_, May 26 2017

%p A016209 := proc(n) (5^(n+2)-2*3^(n+2)+1)/8; end proc: # _R. J. Mathar_, Mar 22 2011

%t Join[{a=1,b=9},Table[c=8*b-15*a+1;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 07 2011 *)

%t CoefficientList[Series[1/((1-x)(1-3x)(1-5x)),{x,0,30}],x] (* or *) LinearRecurrence[ {9,-23,15},{1,9,58},30] (* _Harvey P. Dale_, Feb 20 2020 *)

%o (PARI) a(n)=if(n<0,0,n+=2; (5^n-2*3^n+1)/8)

%o (Magma) [(5^(n+2)-2*3^(n+2)+1)/8: n in [0..20]]; // _Vincenzo Librandi_, Sep 17 2011

%Y Cf. A016218, A016208, A000392, A000225, A003462, A003463, A003464, A023000, A023001, A002452, A002275, A016123, A016125, A016256, A039755, A021424.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_